JavaScript:基于相等的值将数组分割成组(数组数组)

I'm trying to do this exercise:

我试着做这个练习:

The challenge is to implement a function which adds all together all the consecutive numbers in an array and pushes them into a new array. example: sumConsecutives([1,1,2,1,1,1,1,2,1,1,1]); // -> [2,2,4,2,3]

挑战在于实现一个函数，该函数将数组中的所有连续数相加，并将它们推入一个新的数组。例子:sumConsecutives([1,1、2、1,1,1,1,2,- 1,1,1]);/ / - >(2、2、4、2、3)

My idea is to split the array into an array of arrays. So for the above example: [[1,1],[2],[1,1,1,1],[2],[1,1,1]] then go through and reduce them.

我的想法是把数组分割成数组。对于上面的例子:[1,1][2][1,1,1,1][2][1,1,1]然后对它们进行简化。

I've tried with a while loop and push into a temp variable which then gets pushed if the next number is not the same, to no avail.

我尝试了一个while循环，并将它推入一个临时变量，如果下一个数字不相同，它将被推入。

Any ideas on how to achieve this?

关于如何实现这一点，有什么想法吗?

7 个解决方案

#1

The easiest approach I could think of... (without map/reduce, though)

我能想到的最简单的方法是……(虽然没有map / reduce)

var sumConsecutives = function(arr) {
var newArr = [];
var prev = arr[0];
var sum = arr[0];

for (var i=1; i<arr.length; i++){
	if (arr[i] !== prev) {
		newArr[newArr.length] = sum;
		sum = 0;
	}
	sum += arr[i];
	prev = arr[i];
}

// Add last sum
newArr[newArr.length] = sum;

return newArr;
};

console.log ( sumConsecutives([1,1,2,1,1,1,1,2,1,1,1]) );
	

#2

You can use a one step reduce approach:

你可以使用一个步骤减少方法:

const sumConsecutives = ar =>
ar.reduce((ac, x, i) => {
  if ( i !== 0 && ar[i-1] === x)
    ac[ac.length -1] += x;
  else 
    ac.push(x);
    
  return ac;
 }, [])


var r = sumConsecutives([1,1,2,1,1,1,1,2,1,1,1]); // -> [2,2,4,2,3]

console.log(r)

#3

I liked the challange so I made it in a two-step function you thought was useful.

我喜欢challange，所以我用了一个你认为有用的两步函数。

• The first reduce creates the array of arrays

  第一个reduce创建数组的数组。

• The second reduce sums them up.

  第二节总结他们。

It is not the shortest code, but I hope the most understandable.

这不是最短的代码，但我希望最容易理解。

const arr = [1,1,2,1,1,1,1,2,1,1,1];
let currentNumber = undefined;
let currentTempArr = [];

let newArr = arr.reduce((tempArr, value) => {
  // check if current number is set
  if(currentNumber == undefined) currentNumber = value;
  
  // if current number then push to temp array
  if(currentNumber == value)
    currentTempArr.push(value);
  // else just create a new array and push the old one into the parent array
  else {
    tempArr.push(currentTempArr);
    currentTempArr = [];
    currentNumber = value;
    currentTempArr.push(value);
  }
  
  // return the array back to the next reduce iteration
  return tempArr;
}, []);

// push the last temp array, because the function stops before the push
newArr.push(currentTempArr);

// this build your array of arrays
console.info('The array of arrays');
console.log(newArr); // [ [1,1,1], [2], ... ]

// now sum up the sub arrays
let sumArr = newArr.reduce((tempArr, value) => {
  let sum = 0;
  
  // for every value in the array we add that to the sum
  value.forEach(val => sum += val);
  
  // add the total to the temp array
  tempArr.push(sum);
  
  // return the filled array back to the reduce function
  return tempArr;
}, []);

// the array with summed up values
console.info('see the magic happen');
console.log(sumArr);

#4

function sumConsecutives(input) {
    var output = [],
        factor = 1,
        lastnr = null;

    for (var i = 0; i < input.length; i++) {
        if (i === 0) {
            lastnr = input[i];
            continue;
        }

        if (input[i] !== lastnr) {
            output.push(lastnr * factor);
            lastnr = input[i];
            factor = 1;
        } else {
            factor++;
        }

        if (i === (input.length - 1)) {
            output.push(input[i] * factor);
        }
    }
    return output;
}

var result = sumConsecutives([1,1,2,1,1,1,1,2,1,1,1]);

#5

You can use array.reduce

您可以使用array.reduce

Logic

• Create a temp array.
• 创建一个临时数组中。
• loop over array and check if previous and current value.
• 对数组进行循环，检查之前的值和当前值。
• If same, take last element of temp array and add current value to it.
• 如果相同，取temp数组的最后一个元素并向其添加当前值。
• If not, push current element to temp array
• 如果不是，则将当前元素推到临时数组

Sample

function sumConsecutives(arr) {
  var r = [];
  arr.reduce(function(p, c, i, a) {
    if (p === c) {
      r[r.length - 1] += c
    } else {
      r.push(c)
    }
    return c;
  }, 0);
  return r;
}
var a = [1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1];
console.log(sumConsecutives(a));

#6

This is the easiest i could think of:

这是我能想到的最简单的方法:

var testarr = [1,1,2,1,1,1,1,2,1,1,1];
var resultarr = [testarr[0]];
for(var i=1; i<testarr.length; i++){
	if(testarr[i-1]==testarr[i]){
		resultarr[resultarr.length - 1] += testarr[i];
	}else{
		resultarr.push(testarr[i]);
	}
}
console.log(resultarr);

#7

You may also try this approach :

你也可以尝试以下方法:

var source = [1,1,2,1,1,1,1,2,1,1,1];

source.reduce(function(p, c, index, arr) {                 
   if (p.length === 0 || arr[index - 1] !== c) {
       return p.concat(c);
   }
   else if (arr[index - 1] === c) {           
        p[p.length - 1] += c             
   }

   return p;
}, []);